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19 tháng 6 2018

Ta có:

1/5×5 < 1/4×5

1/6×6 < 1/5×6

1/7×7 < 1/6×7

.........

1/100×100 < 1/99×100

=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 < 1/4×5 + 1/5×6 + 1/6×7 +.....+ 1/99×100

                                      = 1/4-1/5 + 1/5-1/6 + 1/6-1/7 +......+ 1/99-1/100

                                    = 1/4-1/100 < 1/4  

=> 1/5×5 + 1/6×6+1/7×7 +...+1/100×100<1/4  (1)

Lại có:

1/5×5 > 1/6×7

1/6×6 > 1/7×8

1/7×7 > 1/8×9

........

1/100×100 > 1/101×102

=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 > 1/5×6 + 1/6×7 + 1/7×8  +.....+1/100×101

                                   = 1/5-1/6 + 1/6-1/7 + 1/7-1/8 +.....+ 1/100 - 1/101

                                   = 1/5 - 1/101 > 1/5 - 1/30 = 1/6

=> 1/5×5 + 1/6×6 +1/7×7 +.....+ 1/100×100>1/6 (2)

Từ (1) và (2)

=> 1/6 < 1/5×5 +1/6×6+ 1/7×7 +...+1/100×100<1/4

19 tháng 6 2018

Đặt \(A=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)

Có \(\frac{1}{5.5}< \frac{1}{4.5};\frac{1}{6.6}< \frac{1}{5.6};...;\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)(1)

Lại có :\(\frac{1}{5.5}>\frac{1}{5.6};\frac{1}{6.6}>\frac{1}{6.7};...;\frac{1}{100.100}>\frac{1}{100.101}\)

\(\Rightarrow A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\left(2\right)\)

Từ (1) và (2) \(\RightarrowĐCCM\)

10 tháng 5 2017

Ta thấy:

1/2*2<1/1*2)vì 2*2>1*2).

1/3*3<1/2*3(vì 3*3>2*3).

...

1/8*8<1/7*8(vì 8*8>7*8).

=>1/2*2+1/3*3+1/4*4+...+1/8*8<1/1*2+1/2*3+1/3*4+...+1/7*8.

=>B<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8.

=>B<1-1/8.

=>B<7/8.

Mà 7/8<1.

=>B<1.

Vậy B<1(đpcm).

10 tháng 5 2017

\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(\Rightarrow1-\frac{1}{8}< 1\)

=>B<1

19 tháng 5 2018

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

19 tháng 5 2018

Ta có : 

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\left(đpcm\right)\)

Chúc bạn học tốt !!! 

10 tháng 5 2017

Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)

26 tháng 7 2016

\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)

\(A< \frac{1}{4}-\frac{1}{100}\)

\(A< \frac{6}{25}< \frac{1}{4}\)

27 tháng 7 2016

các bạn giải mau lên

12 tháng 2 2016

Ta xét : \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{20}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)\)

\(=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+....+\frac{1}{20}\)

Vì \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+....+\frac{1}{20}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)

nên \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+....+\frac{1}{20}\) ( đpcm )

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(\Rightarrowđpcm\)

10 tháng 4 2018

free ire